\newproblem{lay:5_3_29}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.3.29}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	The diagonalization of a matrix is not unique. Given the following diagonalization of the matrix $A$
	\begin{center}
		$A=PDP^{-1}$ \\
		$\begin{pmatrix}7 & 2 \\ -4 & 1\end{pmatrix}=\begin{pmatrix}1 & 1 \\ -1 & -2\end{pmatrix}\begin{pmatrix}5 & 0 \\ 0 & 3\end{pmatrix}
			\begin{pmatrix}1 & 1 \\ -1 & -2\end{pmatrix}^{-1}$
	\end{center}
	Now consider a new factorization of the form $A=P_1D_1P_1^{-1}$ with $D_1=\begin{pmatrix}3 & 0 \\ 0 & 5\end{pmatrix}$. Find the matrix $P_1$.
}{
   % Solution
	$P_1$ is simply the reorganization of the columns in $P$ such that each eigenvector is in the same column as its corresponding eigenvalue in $D_1$
	\begin{center}
		$P_1=\begin{pmatrix}1 & 1 \\ -2 & -1\end{pmatrix}$
	\end{center}
}
\useproblem{lay:5_3_29}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
